\(Q=\dfrac{\left(x-1\right)}{\left(x+1\right)}=\dfrac{x+1-2}{x+1}=\dfrac{x+1}{x+1}-\dfrac{2}{x+1}=1-\dfrac{2}{x+1}\)
Để \(Q\in Z\) \(\Rightarrow\left(x+1\right)\inƯ\left(2\right)\)
\(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
x | -2 | -1 | 1 | 2 |
x+1 | -3 | -2 | 0 | 1 |
Vậy để \(Q\in Z\Rightarrow x=\left\{-3;-2;0;1\right\}\)